I was always amazed that the Black-Scholes delta i.e. the following expression: $$\frac{\partial}{\partial S}\left[ S\cdot \Phi\left(\frac{log \left(\frac{S}{K}\right)+(r+\frac{\sigma^2}{2})T}{\sigma \sqrt{T}} \right)-Ke^{-rT} \cdot \Phi\left(\frac{log \left(\frac{S}{K}\right)-(r+\frac{\sigma^2}{2})T}{\sigma \sqrt{T}} \right)\right]$$ Is just equal to the value of the CDF on the left: $$\Phi\left(\frac{log \left(\frac{S}{K}\right)+(r+\frac{\sigma^2}{2})T}{\sigma \sqrt{T}} \right)$$ And similarly for derivative with respect to $K$ despite the fact that $S$ and $K$ appear in the middle of both of them. Recently I've stumbled upon a paper which states that this fact follows from Euler's theorem as the expression is homogeneous of degree $1$ as a function of $S$ and $K$ and therefore can be written as a linear combination of its partial derivatives: $$f(S,K)=S \cdot \frac{\partial f}{\partial S}+K \cdot \frac{\partial f}{\partial K}$$ I see how this makes the above result much more reasonable but I don't see how it actually proves this fact. Imagine for example the function $f(S,K)=S+K$. We have: $$f(S,K)=S\cdot \frac{K}{S}+K\cdot \frac{S}{K}$$ And yet $\frac{\partial f}{\partial S}=1 \neq \frac{K}{S}$ as well as $\frac{\partial f}{\partial K}=1 \neq \frac{S}{K}$. Can this reasoning be fixed somehow, so that it is possible to compute this derivative without angaging in messy calculations?

  • $\begingroup$ To be honest it is not that hard to compute the derivatives directly as shown here. This is more of a curiosity than anything else. $\endgroup$ – RRL May 18 at 7:47

You just showed that if $f$ is degree-one homogeneous then there exist representations of the form

$$f(S,K) = Sg(S,K) + K h(S,K),$$

where $g$ and $h$ are not the partial derivatives.

The actual statement of Euler's theorem is that a function $f:\mathbb{R}^n\to \mathbb{R}$ with continuous partial derivatives is homogeneous of degree $\alpha$ if and only if

$$\alpha f(x_1, \ldots,x_n) = \sum_{j=1}^n x_j \frac{\partial f}{\partial x_j}$$

It does not say given some representation $ \alpha f(x_1, \ldots,x_n) = \sum_{j=1}^n x_jg_j(x_1, \ldots,x_n)$ and the fact that $f$ is homogeneous of degree $\alpha$, then it must hold that functions $g_j$ are the partial derivatives.

Where this helps with finding the Black-Scholes delta, though, is given that the option value is degree-one homogeneous, we know there exists a representation of the form

$$f(S,k) = S \frac{\partial f}{\partial S} + K \frac{\partial f}{\partial K} = S \Phi(d_1) +K \left(-e^{-rT}\Phi(d_2) \right)$$

By your reasoning, we either have

$$\frac{\partial f}{\partial S} = \Phi(d_1) \quad \text{or} \quad \frac{\partial f}{\partial S} = - \frac{K}{S}e^{-rT} \Phi(d_2)$$

Now consider that for fixed $K$ and $S \to \infty$, the option price behaves asymptotically as

$$f(S,K) \sim S - e^{-rT}K, \quad \frac{\partial f}{\partial S} \sim 1$$

Since, $\Phi(d_1), \Phi(d_2) \to 1$ as $S \to \infty$, only the first representation

$$\frac{\partial f}{\partial S} = \Phi(d_1) \sim 1,$$

has the correct asymptotic behavior.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.